Question-79121

Question Number 79121 by M±th+et£s last updated on 22/Jan/20

Answered by mr W last updated on 23/Jan/20

Commented by mr W last updated on 23/Jan/20

R=radius of semicircle 2a=side length of equilateral triangle λ=(a/R) ((sin (π/3))/R)=((sin (π−β−(π/3)))/a)=((cos (β−(π/6)))/a) cos (β−(π/6))=(((√3)a)/(2R))=(((√3)λ)/2) ⇒β=cos^(−1) (((√3)λ)/2)+(π/6) sin β=(1/2)×(((√3)λ)/2)+((√3)/2)×((√(4−3λ^2 ))/2)=(((√3)(λ+(√(4−3λ^2 ))))/4) α=(π/2)−β=(π/3)−cos^(−1) (((√3)λ)/2) A=2(((αR^2 )/2)+((Ra sin β)/2)) A=R^2 (α+λ sin β) area of triangle =(√3)a^2 =(√3)λ^2 R^2 =A+S ..(i) area of semicircle =((πR^2 )/2)=A+2S ..(ii) 2×(i)−(ii): ⇒2(√3)λ^2 R^2 −((πR^2 )/2)=A=R^2 (α+λ sin β) ⇒2(√3)λ^2 −(π/2)=α+λ sin β ⇒2(√3)λ^2 −(π/2)=(π/3)−cos^(−1) (((√3)λ)/2)+((λ(√3)(λ+(√(4−3λ^2 ))))/4) ⇒7(√3)λ^2 −λ(√(3(4−3λ^2 )))+4 cos^(−1) (((√3)λ)/2)=((10π)/3) ⇒λ≈0.894034 from (ii): S=(1/2)(((πR^2 )/2)−A)=((πR^2 )/4)−(A/2) (S/A)=((πR^2 )/(4×(2(√3)λ^2 R^2 −((πR^2 )/2))))−(1/2) ⇒(S/A)=(π/(8(√3)λ^2 −2π))−(1/2)≈0.155564

$${R}={radius}\:{of}\:{semicircle} \\ $$$$\mathrm{2}{a}={side}\:{length}\:{of}\:{equilateral}\:{triangle} \\ $$$$\lambda=\frac{{a}}{{R}} \\ $$$$\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{3}}}{{R}}=\frac{\mathrm{sin}\:\left(\pi−\beta−\frac{\pi}{\mathrm{3}}\right)}{{a}}=\frac{\mathrm{cos}\:\left(\beta−\frac{\pi}{\mathrm{6}}\right)}{{a}} \\ $$$$\mathrm{cos}\:\left(\beta−\frac{\pi}{\mathrm{6}}\right)=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}{R}}=\frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}} \\ $$$$\Rightarrow\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}}+\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{sin}\:\beta=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\sqrt{\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} }}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}\left(\lambda+\sqrt{\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} }\right)}{\mathrm{4}} \\ $$$$\alpha=\frac{\pi}{\mathrm{2}}−\beta=\frac{\pi}{\mathrm{3}}−\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}} \\ $$$${A}=\mathrm{2}\left(\frac{\alpha{R}^{\mathrm{2}} }{\mathrm{2}}+\frac{{Ra}\:\mathrm{sin}\:\beta}{\mathrm{2}}\right) \\ $$$${A}={R}^{\mathrm{2}} \left(\alpha+\lambda\:\mathrm{sin}\:\beta\right) \\ $$$${area}\:{of}\:{triangle}\:=\sqrt{\mathrm{3}}{a}^{\mathrm{2}} =\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} {R}^{\mathrm{2}} ={A}+{S}\:\:..\left({i}\right) \\ $$$${area}\:{of}\:{semicircle}\:=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}={A}+\mathrm{2}{S}\:\:..\left({ii}\right) \\ $$$$\mathrm{2}×\left({i}\right)−\left({ii}\right): \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} {R}^{\mathrm{2}} −\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}={A}={R}^{\mathrm{2}} \left(\alpha+\lambda\:\mathrm{sin}\:\beta\right) \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} −\frac{\pi}{\mathrm{2}}=\alpha+\lambda\:\mathrm{sin}\:\beta \\ $$$$\Rightarrow\mathrm{2}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} −\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{3}}−\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}}+\frac{\lambda\sqrt{\mathrm{3}}\left(\lambda+\sqrt{\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} }\right)}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{7}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} −\lambda\sqrt{\mathrm{3}\left(\mathrm{4}−\mathrm{3}\lambda^{\mathrm{2}} \right)}+\mathrm{4}\:\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\lambda}{\mathrm{2}}=\frac{\mathrm{10}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{894034} \\ $$$$ \\ $$$${from}\:\left({ii}\right): \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}−{A}\right)=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\frac{{A}}{\mathrm{2}} \\ $$$$\frac{{S}}{{A}}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}×\left(\mathrm{2}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} {R}^{\mathrm{2}} −\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}\right)}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{S}}{{A}}=\frac{\pi}{\mathrm{8}\sqrt{\mathrm{3}}\lambda^{\mathrm{2}} −\mathrm{2}\pi}−\frac{\mathrm{1}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{155564} \\ $$

Commented by mr W last updated on 23/Jan/20